Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e2 + 10;
const int INF = 0x7f7f7f7f;
int n, m;
int grid[MAXN][MAXN];
void oprow(vector<int>& rv) {
  for (int rr = 0; rr < n; rr++) {
    int mmin = INF;
    for (int kk = 0; kk < m; kk++) {
      mmin = min(mmin, grid[rr][kk]);
    }
    if (mmin == 0) continue;
    for (int kk = 0; kk < m; kk++) {
      grid[rr][kk] -= mmin;
    }
    while (mmin--) rv.push_back(rr);
  }
}
void opcol(vector<int>& cv) {
  for (int cc = 0; cc < m; cc++) {
    int mmin = INF;
    for (int kk = 0; kk < n; kk++) {
      mmin = min(mmin, grid[kk][cc]);
    }
    if (mmin == 0) continue;
    for (int kk = 0; kk < n; kk++) {
      grid[kk][cc] -= mmin;
    }
    while (mmin--) cv.push_back(cc);
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  bool zero = true;
  for (int ii = 0; ii < n; ii++) {
    for (int jj = 0; jj < m; jj++) {
      cin >> grid[ii][jj];
    }
  }
  vector<int> rv, cv;
  if (n <= m) {
    oprow(rv);
    opcol(cv);
  } else {
    opcol(cv);
    oprow(rv);
  }
  bool ok = true;
  for (int ii = 0; ii < n; ii++) {
    for (int jj = 0; jj < m; jj++) {
      if (grid[ii][jj] != 0) ok = false;
    }
  }
  if (!ok)
    cout << -1 << endl;
  else {
    cout << rv.size() + cv.size() << endl;
    for (auto v : rv) {
      cout << "row"
           << " " << v + 1 << endl;
    }
    for (auto v : cv) {
      cout << "col"
           << " " << v + 1 << endl;
    }
  }
}